Problem: $ B = \left[\begin{array}{rrr}3 & -2 & 4 \\ 1 & 3 & -1\end{array}\right]$ $ D = \left[\begin{array}{rr}-1 & -1 \\ -2 & 2 \\ 3 & 5\end{array}\right]$ What is $ B D$ ?
Answer: Because $ B$ has dimensions $(2\times3)$ and $ D$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B D = \left[\begin{array}{rrr}{3} & {-2} & {4} \\ {1} & {3} & {-1}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{-1} \\ {-2} & \color{#DF0030}{2} \\ {3} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{-1}+{-2}\cdot{-2}+{4}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{-1}+{-2}\cdot{-2}+{4}\cdot{3} & ? \\ {1}\cdot{-1}+{3}\cdot{-2}+{-1}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{-1}+{-2}\cdot{-2}+{4}\cdot{3} & {3}\cdot\color{#DF0030}{-1}+{-2}\cdot\color{#DF0030}{2}+{4}\cdot\color{#DF0030}{5} \\ {1}\cdot{-1}+{3}\cdot{-2}+{-1}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{-1}+{-2}\cdot{-2}+{4}\cdot{3} & {3}\cdot\color{#DF0030}{-1}+{-2}\cdot\color{#DF0030}{2}+{4}\cdot\color{#DF0030}{5} \\ {1}\cdot{-1}+{3}\cdot{-2}+{-1}\cdot{3} & {1}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{2}+{-1}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}13 & 13 \\ -10 & 0\end{array}\right] $